Protons and neutrons are referred to collectively as nucleons. Nucleons interact via the strong nuclear force, and this interaction can't be expressed by any simple equation. The reason is that nucleons are not fundamental particles. They're actually clusters of quarks.
Short range
The low-energy structure of nuclei is amazingly insensitive to the details of the nucleon-nucleon interaction that you pick as an approximation to the actual underlying quark-quark interaction. This is both good and bad. It's good because you don't need to understand very much about the nasty details in order to find out the properties of nuclei, e.g., why nuclei have the sizes and shapes they do. It's bad because it means that you can never infer very much about the interaction simply by observing properties of nuclei. As an example of how insensitive nuclear structure is to the details of the strong nuclear force, clusters of sodium atoms have magic numbers that match up with the first few magic numbers for nuclei; this is because these magic numbers only depend on the short-range nature of the interaction.
Other effects that can be understood based on the short range of the interaction are:
Nuclei act as though they have surface tension (so they resist being deformed).
Nuclei are most stable if they have even numbers of neutrons and even numbers of protons (because then the neutrons and protons can pair off in time-reversed orbits that maximize their spatial overlap).
Nucleons in an open shell tend to couple so as to form the minimum total angular momentum (the opposite of Hund's rules for electrons).
A residual interaction
The short-range nature of the nuclear interaction is very surprising, because the quark-quark force is believed to be roughly independent of distance. What's happening here is that nucleons are color-neutral, just as a hydrogen atom is charge-neutral. Just as hydrogen atoms "shouldn't" interact, nucleons "shouldn't" interact either. The forces between nucleons very nearly cancel out, and likewise the electrical forces between two neutral hydrogen atoms very nearly cancel out. The nonvanishing interaction comes from effects like the polarization of one particle by the other. For this reason, the nucleon-nucleon interaction is referred to as a residual interaction.
Other than its coupling constant and its range, what other features of the nuclear interaction are important for understanding low-energy nuclear structure?
Spin-orbit
There is a spin-orbit interaction, which is much stronger than, and in the opposite direction compared to, the one expected from special relativity alone.
Symmetry between neutrons and protons
The nuclear interaction remains unchanged when we transform neutrons to protons and protons to neutrons. For this reason light nuclei exhibit nearly identical properties if you swap their N and Z. Heavy nuclei don't have this symmetry, which is broken by the electrical interaction.
No qualitative features inferrable from sizes of nuclei
We do not get any clearcut, qualitative information about the interaction based on the observed sizes of nuclei. An extremely broad class of interactions between point particles results in n-body systems that have bound states and finite density. The finite density (i.e., the lack of a total collapse to a point) is essentially a generic result of the Heisenberg uncertainty principle. Only for certain special types of potentials that blow up to $-\infty$ at short ranges can one circumvent this (Lieb 1976).
A variety of models
Because the nucleon-nucleon interaction is a residual interaction, and nucleons are really composite rather than pointlike, the whole notion of a nucleon-nucleon interaction is an approximation, and one can model it in a variety of ways while still producing agreement with the data. In particular, some models have a hard, repulsive core, while others do not,(Chamel 2010, Stone 2006) and both types can reproduce the observed sizes of nuclei. This disproves the common misconception that such a hard core is needed in order to explain the sizes of nuclei.
References
Chamel and Pearson, 2010, "The Skyrme-Hartree-Fock-Bogoliubov method: its application to finite nuclei and neutron-star crusts," http://arxiv.org/abs/1001.5377
Lieb, Rev Mod Phys 48 (1976) 553, available at http://www.pas.rochester.edu/~rajeev/phy246/lieb.pdf
Stone and Reinhard, 2006, "The Skyrme Interaction in finite nuclei and nuclear matter," http://arxiv.org/abs/nucl-th/0607002
DomDoe's answer is the historical answer, but I suspect that mithusengupta123 may really be asking something like:
Given our understanding of the Standard Model of particle physics, how is it that the low-energy physics of hadrons has an approximate isospin symmetry?
1. Isospin acting on quarks
Historically, isospin historically proposed as a symmetry between protons and neutrons. A nucleon is a field $N = (p, n)^T$ that is a doublet under SU(2) isospin. Equivalently, we may define isospin on the light quarks, placing the up and down into an isospin doublet, $q = (u, d)^T$.
By addition of spin, a state with two up quarks and a down quark has net isospin $I^3 = +1/2$, and so this is consistent with the proton being a $+1/2$ state. Similarly with the neutron having isospin $-1/2$.
2. Isospin is not exact
Isospin is not an exact symmetry. It is an approximate symmetry. It is broken by:
The masses of the quarks. The up quark has a different mass than the down quark. Similarly, the proton has a different mass than the neutron.
It is also broken my electromagnetism. The up and down quarks have different electric charge. Similarly, the proton is charged whereas the neutron is neutral.
In this sense, isospin is not a symmetry. It's almost a symmetry. What do we mean by almost? We mean that there is a small, dimensionless parameter that we can expand about. When we're talking about low-energy hadronic physics, this parameter is something like $(m_u - m_d)/\Lambda_\text{QCD}$, where $\Lambda_\text{QCD}$ is the confinement scale (alternatively you could put in the proton mass, which is roughly the same order of magnitude). The other expansion parameter is $\alpha = 1/137$. Both of these expansion parameters are around the percent level.
This means that if we have a result that is true in the exact limit of isospin symmetry, then the actual result in nature is the same thing up to percent-level corrections. Further, we can use techniques like perturbation theory to solve for these corrections order by order.
3. Isospin in practice
How do we use isospin? One simple example are the pions. We know that pions are bound states of two light quarks. That is, they are in an isospin representation that comes from the product of two doublets. (There's a subtlety here because it's really a quark--anti-quark pair, see e.g. this question).) We know that the combination of two SU(2) doublets gives a triplet and a singlet.
Experimentally, we can identify the triplet and the singlet as the three pions and the $\eta$, respectively. The three pions are related to each other by isospin symmetry, whereas the $\eta$ is its own object. Indeed, the $\eta$ is about four times heavier than the pions, which are roughly the same mass.
On the other hand, the pions do not have the same exact mass. The charged pions are 140 MeV, while the neutral pion is 135 MeV. This few percent correction to the exact isospin limit is precisely the result of the mass splitting of the quarks and the electromagnetic discrimination between the charged and neutral states.
4. Where did isospin come from?
Now to the crux of the question: if we know the Standard Model, how do we understand that at low energies there is an approximate isospin symmetry? How is this related to any of the other symmetries of the Standard Model?
The answer is that isospin symmetry is the result of chiral symmetry breaking.
Imagine writing down all of the Standard Model particles without any interactions. There is a symmetry among the quarks. $U(6)_L\times U(6)_R$, which rotates the six left-handed quarks separately from the six right-handed quarks. (Recall from representation theory that left-chiral and right-chiral fields are, a priori, completely different things which can have different charges.) This is actually
This symmetry is broken by:
The electroweak force, which distinguishes between left- and right-chiral quarks. Further, it places left-handed quarks into doublets and distinguishes between right-handed quarks with up-type and down-type charges.
The Yukawa interactions with the Higgs, which (upon electroweak symmetry breaking) gives different masses to vectorlike combinations of left- and right-chiral quarks. This combines Weyl fermions into Dirac fermions.
Let us ignore the electroweak force---these effects come with electroweak couplings which we know are relatively small. Certainly at low energies when they are either mediated by a photon ($\alpha = 1/137$) or a $W/Z$ boson (suppressed contributions at low energies because they're heavy).
Then the mass terms pair left- and right-chiral fermions. These mass terms look like $m_q \bar q_L q_R+\text{h.c.}$. First, let us pretend that all of the quarks have the same mass. That is, $m_q$ is universal. Then this means that our original $U(6)_L\times U(6)_R$ symmetry is broken to $U(6)_D$, where the $D$ means diagonal. If you rotate between the six left-handed quarks, you have to do a compensating rotation among the six right-handed quarks in order for the mass term to remain invariant.
Once you turn on the different masses of each quark, then this $U(6)_D$ is broken further to $U(1)^6$, which is basically the rephasing of each type of quark.
We know that there's a big hierarchy in the quark masses, so for the most part, this breaking down to $U(1)^6$ symmetry is pretty rigorous. Each U(1) represents the conservation of up-ness, down-ness, strangeness, charm-ness, etc. (We know that the interactions of the $W$ boson violate these, but for now we're ignoring the electroweak interactions.) However, the mass splitting between the up and the down quark is relatively modest... so in fact, the up and the down have an approximate $SU(2)$ symmetry left over. (I'm being sloppy with the U(1) factors, you can repackage them into overall baryon number conservation and other conservation laws.) This $SU(2)$ symmetry is precisely isospin.
5. What does this buy us?
Why is this useful from the fundamental point of view?
We know how to deal with spontaneous symmetry breaking. In particular, we know that the breaking of the $SU(2)_L\times SU(2)_R$ subgroup of $U(6)_L\times U(6)_R$ is a spontaneous breaking of an approximate symmetry. Thus we can describe the interactions of the Goldstone bosons of this breaking by the nonlinaer sigma model, up to corrections.
The power of this point of view is that the pions are identified with the Goldstone bosons of $SU(2)_L\times SU(2)_R \to SU(2)_D$. Their interactions with each other are predicted by the non-linear sigma model once one defines the scale at which this symmetry is broken. (This is the pion decay constant, which is related to the chiral symmetry breaking scale by the QCD chiral condensate.)
One can extend this and say that the strange quark is also reasonably close in mass to the up and down. Then one can talk about an approximate $SU(3)_L\times SU(3)_R \to SU(3)_D$ breaking. In the limit where the masses of all three light quarks are degenerate, then one can describe the Goldstone bosons as an octet of pions and kaons. The interactions between these particles are all predicted by the nonlinear sigma model, up to corrections that are now a little bigger than in the pure SU(2) isospin case.
Best Answer
In the standard model of particle physics, there exist elementary particles out of which all other matter is composed.
The proton and the neutron are composed out of up and down quarks
the proton is (u u d) and the neutron (u d d). If you notice in the table the quarks have different charges. In the hypothetical case of no charges, there will still be the symmetry to which the strong interaction ties the quarks, so a proton and a neutron will still be different, occupying the isotopic spin +1/2 and -1/2 respectively in the representation. They will have the same strong force effects as composites of quarks , following the symmetries.
This is just a statement to demonstrate the isotopic spin symmetry that arises from the strong interactions. Charges are fundamental in nature because we cannot turn them off.